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Two gaseous reactants are allowed to react in a 1-L flask, and the reaction rate is measured. The experiment is repeated with the same amount of each reactant and at the same temperature in a 2-L flask (so the concentration of each reactant is less). What is likely to happen to the measured reaction rate in the second experiment compared to the first?

Short Answer

Expert verified
The measured reaction rate in the 2-L flask is likely to decrease due to the lower concentration of reactants.

Step by step solution

01

- Understanding Reaction Rates

Reaction rates are typically proportional to the concentration of the reactants. This is because a higher concentration means more reactant molecules are in close proximity, thus increasing the chances of collision and reaction.
02

- Comparing Concentrations in Different Flask Sizes

In a 2-L flask, the concentration of the gaseous reactants is halved because the same amount of gas that originally occupied 1-L now occupies 2-L. This reduces the number of reactant molecules per unit volume.
03

- Predicting the Effect on Reaction Rate

With a reduced concentration in the 2-L flask, the reaction rate is expected to decrease because there will be fewer collisions between reactant molecules. For many reactions, the rate is directly proportional to the concentration of reactants, assuming other conditions remain the same.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Kinetics
Chemical kinetics is the study of reaction rates and the factors that affect them. It helps us understand how different conditions influence the speed at which chemical reactions occur. Factors such as temperature, concentration, surface area, and the presence of catalysts can significantly impact the rate of a reaction.

By examining these factors, chemists can optimize reactions for industrial processes, predict reaction behavior, and control product formation. Chemical kinetics is fundamental in fields ranging from pharmaceuticals to environmental science, allowing us to design and manage chemical processes efficiently.
Reaction Rates
Reaction rates express the speed at which reactants are converted into products in a chemical reaction. They can be measured by the change in concentration of reactants or products per unit time. Higher reaction rates signify a faster reaction process.

A reaction that takes place quickly, like an explosion, has a high rate, while one that occurs slowly, such as rust formation, has a low rate. Understanding and controlling reaction rates is crucial for safety and economic reasons, especially in industrial chemical processes.
Effect of Concentration on Reaction Rate
The concentration of reactants is a major factor affecting the rate of a chemical reaction. Generally, a higher concentration of reactants leads to a higher reaction rate. This is due to the increased number of molecules or ions in a given volume, which raises the probability of collisions and successful reactions between them.

When the concentration decreases, the frequency of collisions drops, and so does the reaction rate. This relationship is often described by the rate law for a reaction, where the rate is proportional to the concentration of reactants raised to a power, known as the order of the reaction.
Comparing Flask Volumes in Reactions
When comparing reactions in flasks of different volumes, the concentration of reactants changes according to the volume they occupy. If the amount of reactant and the temperature are kept constant, but the volume of the flask is doubled, the concentration of the reactant is halved.

This lower concentration in the larger flask means that reactant molecules are more spread out, reducing the likelihood of collisions and, consequently, the reaction rate. Understanding the impact of flask volume on reaction concentration is essential in experimental design and interpreting results in chemical kinetics.

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Most popular questions from this chapter

This reaction is exothermic. $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \rightleftharpoons 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ Predict the effect (shift right, shift left, or no effect) of these changes. (a) increasing the reaction temperature (b) decreasing the reaction temperature

Why are solids and liquids omitted from the equilibrium expression?

Write an equilibrium expression for each chemical equation involving one or more solid or liquid reactants or products. (a) \(\mathrm{HCHO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CHO}_{2}^{-}(a q)\) (b) \(\mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q)\) (c) \(2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\) (d) \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\)

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This reaction is exothermic. $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}(g) $$ If you were a chemist trying to maximize the amount of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) produced, which of the following might you try? Assume that the reaction mixture reaches equilibrium. (a) increasing the reaction volume (b) removing \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) from the reaction mixture as it forms (c) lowering the reaction temperature (d) adding \(\mathrm{Cl}_{2}\)

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